codeforces Educational Codeforces Round 33 (Rated for Div. 2)B

B. Beautiful Divisors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently Luba learned about a special kind of numbers that she calls beautiful numbers. The number is called beautiful iff its binary representation consists of k + 1 consecutive ones, and then k consecutive zeroes.

Some examples of beautiful numbers:

• 12 (110);
• 1102 (610);
• 11110002 (12010);
• 1111100002 (49610).

More formally, the number is beautiful iff there exists some positive integer k such that the number is equal to (2k - 1) * (2k - 1).

Luba has got an integer number n, and she wants to find its greatest beautiful divisor. Help her to find it!

Input

The only line of input contains one number n (1 ≤ n ≤ 105) — the number Luba has got.

Output

Output one number — the greatest beautiful divisor of Luba's number. It is obvious that the answer always exists.

Examples

input

3

output

1

input

992

output

496

mad  留下了 不细心看题 代码能力垃圾啊的眼泪    刚开始没看清题 写的很麻烦 之后没又加 等号 一直wa 

#include<bits/stdc++.h>using namespace std;long long xxx(string x){   int a[10001];   for(int j=0;j<x.size();j++){     a[j]=x[j]-'0';   }   long long l=0;   long long i=x.size();   for(int j=0;j<x.size();j++){     l+=pow(2,i-j-1)*a[j];   }   return l;}int main(){   int n;     cin>>n;   string a;   int i=0;   int b[10001];   for(int j=1;;j++){      for(int k=0;k<j;k++){         a=a+"1";      }      for(int k=0;k<j-1;k++){         a=a+"0";      }      b[j]=xxx(a);      if(b[j]>10000000){        i=j;        break;      }      a="";   }   int x=0;   for(int j=1;j<=i;j++){      if(b[j]<=n&&n%b[j]==0){  //这里忘记加了          x=b[j];      }      else if(b[j]>n) break;   }   cout<<x<<endl;   return 0;}